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\title{Superconductivity - Assignment 5}
\author{
	Kees van Kempen (s4853628)\\
	\texttt{k.vankempen@student.science.ru.nl}
}
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\section{$T_c$ upper limit in BCS}
In BCS theory, the formation of Cooper pairs is mediated by phonons.
There is a phonon-electron interaction quantified by the dimensionless quantity 
\[
	\lambda := Vg(\epsilon_F)
\]
with $V$ Cooper's approximate potential and $g(\epsilon_F)$ the density of states near the Fermi surface for the electrons.
A thorough discussion can be found in Annett's book \cite[chapter 6]{annett} and in the slides of week 6 of this course.
The binding energy of the Cooper pairs (i.e. the energy gain of forming these pairs) is
\[
	 -E = 2\hbar\omega_De^{-1/\lambda} =: \Delta_0,
\]
which is also called the gap parameter $\Delta_0$ at zero temperature for BCS.
In the weak coupling limit of the BCS theory, the case we have considered so far, it is assumed that $\lambda << 1$.
It should be noted that this weak limit also means that the gap is smaller than the thermal energy of the highest excited energy phonon, which corresponds to the Debye temperature
\[
	\Delta < k_B\Theta_D.
\]
It is when this assumption breaks down, BCS does not work and we find an upper limit to the critical temperature $T_c$.
We will look at a way to express the critical temperature in terms we can derive, and then look at the values that maximize this critical temperature whilst still following BCS theory.

From the derivation of the BCS coherent state, this gap parameter at finite temperature is found.
There is a temperature dependence $\Delta(T)$ as in figure \ref{fig:gap-T}.
\begin{figure}
	\centering
	\includegraphics[width=.4\textwidth]{Lecture-7-slides-for-printing-slide-13-gap-parameter.pdf}
	\caption{By taking the gap parameter to zero, we find the critical temperature. Figure from the slides of lecture 7.}
	\label{fig:gap-T}
\end{figure}
For larger temperatures, thermal energy is increased, and less energy is required to break up Cooper pairs, thus degrading the superconductivity.
This puts a limit $T_c$.

We will mostly follow the derivation by Waldram \cite[paragraph 7.9, mostly p.128--130]{waldram}.
The superconducting state breaks down at high temperature, at which also $\Delta$ vanishes so that the gap parameter is a good order parameter for the state.

Let's consider the gap parameter
\[
	\Delta_{\vec{k}} = -\sum_{\vec{k'}}(1-2f_{\vec{k'}})u_{\vec{k'}}v_{\vec{k}}V_{\vec{k'}\vec{k}},
\]
with $u$ and $v$ occupation functions for the BCS state, $f$ the Fermi occupation number, and $V$ the potential between the states.
Minimizing $\Delta_{\vec{k}}$ and taking that $V_{\vec{k'}\vec{k}} = -V$ is constant gives us a self-consistent relation for the gap parameter.
We also recognize that the states that we sum over all all those states such that they have smaller energy than the highest excited phonon.
\[
	\Delta_{\vec{k}} = V\sum_{\epsilon_{\vec{k'}}}(1-2f_{\vec{k'}})\frac{\Delta_{\vec{k'}}}{2E_{\vec{k'}}}.
\]
Now the right-hand side is independent of $\vec{k}$ but does contain $\Delta_{\vec{k'}}$.
We can thus conclude that the gap parameter should be constant over all states $\vec{k}$!
That means we can divide both sides by it, giving us
\[
	1 = V\sum_{\epsilon_{\vec{k'}}}(1-2f_{\vec{k'}})\frac{1}{2E_{\vec{k'}}}.
\]
Converting the equation to an integral, and substituting in $f(E) = [\exp{(E/(k_BT))}+1]^{-1}$ and $E = \sqrt{\epsilon^2 + \Delta(T)^2}$ yields
\[
	1 = 2g(\epsilon_F)V\int_0^{k_B\Theta_D}\frac{1-2[\exp{(E/(k_BT))}+1]^{-1}}{2\sqrt{\epsilon^2 + \Delta(T)^2}} \textup{d}\epsilon.
\]

I believe Waldram that one could find that
\[
	T_c = 1.14\Theta_D\exp{(-1/(g(\epsilon_F)V))} = 1.14\Theta_D\exp{(-1/(\lambda)}
\]
from this nice equation.

As limiting value, we take $\lambda = 0.3$, as was posed as a reasonable limit for the weak coupling by Alix in lecture 7,
although Waldram \cite{waldram} thinks it is more like $\lambda \approx 0.4$.

For metals, Waldram thinks $\Theta_D \leq \SI{300}{\kelvin}$ is a good limit.
This leads to our final maximum
\[
	T_c \leq 1.14 \cdot 300 \cdot \exp{(-1/0.4)} \approx \SI{28}{\kelvin}.
	%https://www.wolframalpha.com/input?i=1.14*300*e%5E%28-1%2F.4%29
\]
(Using $\lambda = 0.3$ yields an even lower $T_c \leq \SI{12}{\kelvin}$.)

For larger $T_c$ values, larger binding of Cooper pairs would be needed to overcome the thermal energy.
This means our assumption of weak coupling breaks down, making most of the derivation invalid without further arguments.

\section{Energy gap $\Delta$ et al.}


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